What will happen if a moon sized object collides with Earth?

[Braznor]

The scenario is about a moon sized object colliding with a planet similarly sized to Earth. What I need is a rough idea about the moon's velocity needed so as to shatter the planet completely. The scenario calls for the total dispersal of the planetary debris into space without giving them the chance of collude themselves back into a planet, so I would need the velocity to frame the situation around it.

Can someone help me with the maths to find it?

 

[AnnonUSA]

Can't help with the math, but that would indeed be a really bad hair day.....

 

[Peter]

You'll also need to account for elasticity. Earth is surprisingly liquid except for the bit of surface we're sitting on.

 

[dkozloski]

Visualize poop hitting the fan.

 

[BrownTown]

The real question is whether or not the people reading this will actually know what the velocity would be themselves? If not just make up a number and thats good enough.

 

[CycloWizard]

Originally posted by: Peter
You'll also need to account for elasticity. Earth is surprisingly liquid except for the bit of surface we're sitting on.

There are solutions for a linearly elastic sphere filled with a purely viscous Newtonian liquid. Of course, this would be a gross oversimplification in this case, but it would be an interesting place to start. The article that I recall is regarding parallel plate compression of an alginate hydrogel containing water and may be written by Tatara, though I havent' read it for a while and could be mistaken. In any case, this solution would not tell you anything about the fracture characteristics of the planet.

If you really want to do a back-of-the-envelope calculation, you could compute the total energy of the earth (E=m*c^2 ). You would then compute the velocity v required to achieve that energy by a body with the same mass as the moon (E=1/2*m*v^2). This velocity would be much larger than necessary, but would give you a basic idea anyway.

 

[CTho9305]

Originally posted by: CycloWizard
If you really want to do a back-of-the-envelope calculation, you could compute the total energy of the earth (E=m*c^2 ). You would then compute the velocity v required to achieve that energy by a body with the same mass as the moon (E=1/2*m*v^2). This velocity would be much larger than necessary, but would give you a basic idea anyway.

That's kind of ridiculous. You could get a much lower upper bound by taking the combined mass of the earth and the colliding object, computing the escape velocity (using a simplified equation - when the two masses are similar it's uglier), and the energy to accelerate the mass to that velocity. I came up with 4*10^32 joules, or a nuclear power plant running for 10 billion years.

 

[Lemon law]

Look at it on the bright side, if an earth moon sized object hit the earth, you would not need to write any such paper.

 

[CycloWizard]

Originally posted by: CTho9305
That's kind of ridiculous. You could get a much lower upper bound by taking the combined mass of the earth and the colliding object, computing the escape velocity (using a simplified equation - when the two masses are similar it's uglier), and the energy to accelerate the mass to that velocity. I came up with 4*10^32 joules, or a nuclear power plant running for 10 billion years.

Yes, it is ridiculous. I thought that was the point. We actually had this discussion a while ago, but I'm too lazy to dig through many pages to find the thread. Your method would probably give a closer number, but it would be far too low to really achieve the destruction of the earth due to viscous damping effects since it assumes a perfectly elastic failure mechanism. Mine would grossly overestimate it. Perhaps solving the two problems and taking the geometric mean of the results would give a better approximation.

 

[oynaz]

I think the guys at http://www.badastronomy.com love to do these kind of calculations.

 

[Soccerman06]

link

That is basically what would happen if an object the size and mass of the moon collided with the Earth, sorry I do not know the speed. Although, it takes a massive amount of energy to "split" the earth, and probably a neutron star moving at an extreme velocity just to cut through the earth because of the extreme temperatures *so that it would not melt the object at impact) and mass of the earth.

 

[Braznor]

Okay, I think I have a fair idea what to do. I'm gonna make the velocity a few kilometers short of the speed of light just for the heck of it!

What happens is that a mammoth asteroid gets sucked upwards into unfolded Calabi Yau space and returns back to normal space time right on collision course with the planet.

Would the resulting explosion destroy the solar system itself?

 

[DrPizza]

Don't forget, if it's moving at nearly the speed of light directly toward us (or rather, directly toward where we're going to be at the time it gets here), then we're not going to see it until it's much too late. Think of how hard it was to find the sub-planets in the Kuiper belt.. And, they're barely moving! (relatively speaking.)


Anyone remember the fun thread about covering the moon in aluminum foil?

 

[Nathelion]

Covering the moon in aluminum foil? lol, sound awesome. Must have slipped past me completely though.

If your asteroid is travelling at, say, 2.9*10^8 m/s, then yes, the energy release would probably obliterate the solar system if the thing has any respectable mass. I'll think about it and calculate a bit and then post again with what I come up with...

 

[Nathelion]

Some quick calculations reveal that a moon-sized object traveling at .9c would carry something like 2.55*10^39 J of kinetic energy. For comparison, the sun outputs ~3.864*10^26 W, so it would take 2.09*10^5, or 209 000, years for it to produce this much energy.
According to this www.as.utexas.edu/astronomy/education/spring06/komatsu/secure/lecture05.pdf the gravitational binding energy of the sun is approx. 2.3*10^41 J, so our projectile would definitely not obliterate the solar system. For that, I had to push the speed up to something like .999 c to reach the same order of magnitude. It'd still be one h*ll of a bang, and I imagine that there wouldn't be much left in terms of planets, asteroids, and oort cloud left after our mammoth asteroid is done with the place.

But before we get ahead of ourselves, according to http://www.nmm.ac.uk/server/show/ConWebDoc.730 the energy release in a type Ia supernova is on the order of 2.5*10^46 J, so we still have a ways to go before it gets really violent.